Electromagnetism - Polarization

Questions 54-55 concern a plane electromagnetic wave that is a superposition of two independent orthogonal plane waves and can be written as the real part of 

E = x̂ E₁ exp [i(kz −ωt)] +  ŷ E₂ exp [i(kz − ωt + π)]

where k, ω, E₁ and E₂ are real

If the plane wave is split and recombined on a screen after the two portions, which are polarized in the x- and y- directions, have traveled an optical path difference of 2π/k, the observed average intensity will be proportional to

A. E₁² + E₂² 
B. E₁² − E₂² 
C. (E₁+ E₂)² 
D. (E₁− E₂)² 
E. 0

(GR9677 #55)

Solution:

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i(kz − ωt +π )} 

Path difference of 2π/k
E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[k(z + 2π/k) − ωt + π)} 

E = x̂ E₁ e^{i(kz − ωt)}  +  ŷ E₂  e^{i[kz − ωt)} · e^i3π 
e^i3π = −1

E = x̂ E₁ e^{i(kz − ωt)}  −  ŷ E₂  e^{i(kz − ωt)} 

Intensity in the x- directions, I_x = |E₁|²  
Intensity in the x- directions, I_y = |−E₂|² 

I_total = I_x + I_y = E₁² + E₂²

Answer: A

Optics - Polarization

Unpolarized light of intensity I₀ is incident on a series of three polarizing filters. The axis of the second filter is oriented at 45^o to that of the first filter, while the axis of the third filter is oriented at 90^o to that of the first filter. What is the intensity of the light transmitted through the third filter?

A. 0
B. I₀/8
C. I₀/4
D. I₀/2
E. I₀/√2

(GR0177 #51)

Solution:

Polarizer:
The first filter always reduces the intensity of the light to half, I₁=½ I₀
The next filter reduces the intensity by I_n = I_(n-1) cos² θ
θ is the angle with respect to the n^th filter.

Thus,

I₁ = ½ I₀

I₂ = I₁ (cos 45)² = I₁ (½ √2)² = ½ I₁ = ¼ I₀

I₃=I₂ (cos 45)² = ½ I₂ = I₀/8

Answer: B